Week 5, Operational Amplifiers I

Week 5, Operational Amplifiers I

Summary: Simulate a signal amplification circuit with an op amp. The adjustable signal voltage is replaced by a resistor voltage divider. The output of the op amp is the amplified signal output. The gain is designed to be 10 and all components should satisfy their power limit.

Schematic. The 5V power supply is replaced by one of the 12V.

Calculation for devising the circuit.

 

Components used for the circuit.

 

 

Final setup.

Collected data. The gain is close to 10.

 

 

Measure the power supply current to check if they satisfy the designed current limit. Because I(v1) support both the op amp and the circuit, it is much large than I(v2).

Follow up calculation.

Notes:

1. Because only 1/8W resistors are available, all resistor are designed to satisfy 1/8W limit. Also, because 1k resistors were not available, the Ri and Rf were 2.2k/22k rather than 1k/10k.
2. I measured the total current of power supply one, while I was supposed to measure the current going into the op amp. The KCL was not able to be verify.
3. The gain is about 5% to 10% different to 10.
4. If Rx and Ry are smaller, the result should be better because the amplification circuit has less effect on the series voltage divider.

Week 4, Thevenin Equivalents

Week 4, Thevenin Equivalents

Summary: To compare the output of a linear circuit and its Thevenin equivalent circuit.
1. Build the Thevenin equivalent circuit and measure the voltage cross the terminals of a load resistor.
2. Build the original linear circuit and measure the same voltage. Also, change the load with a factor of R(Th) and verify the load.

The schematic of the original circuit.

The schematic of the Thevenin equivalent circuit.

Calculate the theoretical open circuit voltage, expected voltage with  minimum load(8 V), V(TH), R(TH), for devising the equivalent circuit and comparing the measured values and theoretical values.

Set up the Thevenin equivalent circuit and measure voltage with different load.

Set up the original circuit and do the same measurement. Calculate the power of different load.

Notes:
1. Comparing the Thevenin equivalent circuit and the original circuit, both the open circuit voltage and the voltage with load are very closed, indicating that they are actually equivalent.
2. When R(L)=R(TH), the power of the load is maximum. The result from the last chart agrees this theorem.

Week 3, Transistor Switching

Week 3, Transistor Switching

Summary:
1. Setup a simple circuit with a transistor, switching base current on and off and check the result on the emitter side. Then, use finger tip as the base resistor and try again.
2. Collect current reading from base current and emitter current. The result is expected to be a linear relationship.

Schematic:

Final setup. R1 is the resistor box.

Switching the base current on and off by touching wires. Then replace R1 with finger tip.

Schematic for finding the relationship between base current and emitter current.

Setup:

Data.

Graph. The last point is not used because it is far away from the line.

Notes:
1. The base is the "switch" for the emitter. If there is no current go in the base, the "switch" is off.
2. If the base current if within a certain range, the emitter current is in proportional with the base current. The graph above shows this property; however, the received beta value is far away from what it is supposed to be.

Week 3, Nodal Analysis

Week 3, Nodal Analysis

Summary: Use nodal analysis to find the voltage of a circuit. Construct the circuit, measure voltage and currents, and compare the measured values with the theoretical values.

Schematic
         

Equations for nodal analysis. The theoretical values for V2 and V3 are at the bottom.

Other theoretical calculation for the circuit.

 The actual values of the resistors and DC sources.

Final setup. The positions of all components are the same as they are in the schematic.

Measured values. Except for I(Bat2), all other values are very close to their theoretical values. See notes for the big percent error of I2.

If V2=V3=9V, plug in the equations above one gets V(Bat1)=9.9V, V(Bat2)=10.98V. Using 9.9V and 11.0V, the measured results are very close to the calculated results.

Notes: The reasons for the big difference of I(Bat2) may include:

1. The actual values are slightly different from the desired values. The calculated value will be 2.06mA if one uses all the actual values. The percent of error will be within a reasonable range.
2. The 1.5mA is a very small number. Usually meters are not as accurate in small scale as in large scale.

Week 2, Introduction to Biasing

Week 2, Introduction to Biasing

Summary: In a parallel circuit, different branches share the same power supply; however, the components in each branch may only work under the supply voltage. Resistors are used to drop the voltage in a branch in order to create proper voltage for the load components.

Schemetic

Designed values. Because resistors are only availabe in certain values, the actual R1 is 220 ohms, and R2 is 470 ohms.

Final setup.

Connect meters to different branches to collect readings.

Data

a. The operating time for 0.2A-hr.

b. The percent error between the theoretical value and the actual value is
-19.21% . The causes include:

1.  the actual LED working voltage and amperage may not be the same as labled.
2. The LEDs may not follow Ohm's Law perfectly.
3. The meters affected the circuit and slightly affected the reading.

c. Calculate the power efficiency.

d. If the vattery is 6V, efficiency will go up because the LEDs take a larger portion of the voltage while the current stay the same. For the same reason, the lower voltage, the higher efficency. The lowest voltage is 5V in this case.
(5*22.75+2*20)/(5*(22.75+20))=71.9%

Bouns

 Notes: The actual resistors used are much larger than the designed values; therefore, the readings are also very different from the designed values. The LEDs probably do not follow Ohm's Law in its entire working voltage range. Overall the biasing resistors allow both LEDs work in this circuit.

FreeMat Exercise

FreeMat Exercise

Solve for current.

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Plotting
Assignment 1
1.

From this graph, circuit A will have the lowest output sooner.

2.

Assignment 2
1.

Both the sin and cos part have a period of pi, and the amplitude matches calculated value.

2.

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Complex Numbers
1.

 2.

3.

4.

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Solving for roots
1.

2.

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Notes: The polar numbers are represented as separated amplitude and angle. Not sure if it has a combined notation. I am looking at the syntax for functions and try to make the scripts more useful.